Linear instability implies nonlinear instability I am trying to understand the following proof that linear instability implies nonlinear instability. Suppose we have the ODE, $ \frac{du}{dt}=A(u)$ for which $0$ is a solution. Suppose  $L $ is the linearization and there exists an eigenvalue $\lambda$ of $L$ such that $Lv=\lambda v$ where real part of $\lambda > 0$. Suppose $u(0)=\delta v$ and suppose we write $du/dt= Lu + (A(u) - Lu)$. We have the variation of parameters formula as
$$||u(t)-e^{Lt} \delta v|| \leq \int_0^t ||e^{L(t-s)}||A(u(s))-Lu(s)||ds \leq C_\epsilon \int_0^t e^{(\epsilon+ Re \lambda)(t-s)}||u(s)||^2 ds$$ as long as $u(t)$ remains close to $0$. Then the proof I am using says the following:
Show that $e^{Lt}v= e^{\lambda t}v$ dominates the nonlinear term. I am unable to understand what one must do now. 
Thank you. 
Note: the proof is available on page 3 of this document:
http://depts.washington.edu/bdecon/workshop2012/g_stability.pdf
 A: There are several definitions of local stability (orbital and asymptotic essentially, I think). I assume that what you want to show that $0$ is not locally asymptotically stable, that is for some $\eta>0$, there exist a perturbation $\delta v$ arbitrarily small such that the solution of the ODE with initial data $\delta v$ does not remain in the ball $B_{\|\cdot\|}(0,\eta)$ at all times. I also adopt the notation $u\in X$, $X$ being a Banach space. I denote by $\|u\|_{L^\infty([0,t),(X,\|\cdot\|))}\leq C_0$ the fact that the inequality of the question is true only as long as $u(t)$ remains close to $0$.
Here is how I would proceed:
First, I would continue the estimate on $\|u(t)-e^{Lt}\delta v\|$, for $t\geq 0$, and $\|u\|_{L^\infty([0,t),(X,\|\cdot\|))}\leq C_0$:
$$\|u(t)-e^{Lt}\delta v\|\leq \|u\|_{L^\infty([0,t),(X,\|\cdot\|))}^2\int_0^te^{(\epsilon+\mathcal Re\, \lambda)(t-s)}\,ds\leq \frac 2{\mathcal Re\, \lambda}\|u\|_{L^\infty([0,t),(X,\|\cdot\|))}^2,$$
provided that $\varepsilon>0$ is small enough (ie $\epsilon\leq \frac{\mathcal Re\, \lambda}2$).
Then, I notice that $\|e^{Lt}\delta v\|=e^{t\,\mathcal Re\, \lambda}\|\delta v\|$.
Finally, I use the triangular inequality to get for $t\geq 0$:
$$\|e^{Lt}\delta v\|\leq \|u(t)\|+\|u(t)-e^{Lt}\delta v\|,$$
and then, thanks to previous estimates, if $\|u\|_{L^\infty([0,t),(X,\|\cdot\|))}\leq C_0$:
$$e^{t\,\mathcal Re\, \lambda}\|\delta v\|\leq \|u\|_{L^\infty([0,t),(X,\|\cdot\|))}+\frac 2{\mathcal Re\, \lambda}\|u\|_{L^\infty([0,t),(X,\|\cdot\|))}^2,$$
I then chose $\eta:=\frac {C_0}2$. If I assume by contradiction that $\|u(t)\|\leq \eta$ at all times, then the above inequality is true for all $t\geq 0$, and its right term is bounded. This is absurd, since the left term grows exponentially.
